Correct Answer - Option 2 : 3
Concept:
Let us consider sequence a1, a2, a3 …. an is an A.P.
- Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
- nth term of the A.P. is given by an = a + (n – 1) d
- Sum of the first n terms = Sn =\(\rm \frac n 2\) [2a + (n − 1) × d]= \(\rm \frac n 2\)(a + l)
Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term
Calculation:
Let the first term of AP be 'a' and the common difference be 'd'
Given: Fourth term of an A.P. is zero
⇒ a4 = 0
⇒ a + (4 - 1) × d = 0
⇒ a + 3d = 0
∴ a = -3d .... (1)
To Find: \(\rm \dfrac{t_{25}}{t_{11}}\)
\(\rm \Rightarrow \dfrac{t_{25}}{t_{11}} = \dfrac {a+24d}{a+10d}\\=\dfrac {-3d+24d}{-3d+10d}\\=\dfrac {21d} {7d}=3\)